SAT Subject Test Chemistry

PART 3

PRACTICE TESTS

Practice Subject Tests in Chemistry

The Subject Tests in Chemistry are planned to test the principles and concepts drawn from the factual material found largely in inorganic chemistry and, to a much lesser extent, in organic chemistry. Only a few questions are asked concerning industrial or analytical chemistry.

A detailed description of every aspect of the test is given in the introduction of this book. Read it carefully. Study the types of questions asked on the test. Carefully read the instructions given for each type of question at the beginning of each section. Note: The Practice Test questions contain hyperlinks to their answers and explanations. Remember to click on the question numbers to move back and forth.

You will be provided with a Periodic Table to use during the test. All necessary information regarding atomic numbers and atomic masses is given on the chart. Before you attempt any of the practice tests, read the information in the Introduction and the material that precedes the Diagnostic Test in the front of this book. When you understand the information there, are aware of the types of questions and their respective directions, and know how the test will be scored, you are ready to take Practice Test 1.

Remember that you have 1 hour and that you may not use a calculator. Usethe Periodic Table provided for the Diagnostic Test in the front of this book, and record your answers in the appropriate spaces on the answer sheet. After you have taken each test, follow the instructions for diagnosing your strengths and weaknesses and for how to improve in those areas.

All Practice Answer Sheets are for reference only. Please record all of your answers separately.

Answer Sheet

P R A C T I C E T E S T 1

Determine the correct answer for each question. Then, using a No. 2 pencil, blacken completely the oval containing the letter of your choice.

Practice Test 1

Note: For all questions involving solutions and/or chemical equations, assume that the system is in water unless otherwise stated.

Reminder: You may not use a calculator on these tests.

The following symbols have the meanings listed unless otherwise noted.

H = enthalpy

g = gram(s)

M = molar

J = joules(s)

n = number of moles

kJ = kilojoules

P = pressure

L = liter(s)

R = molar gas constant

mL = milliliter(s)

S = entropy

mm = millimeter(s)

T = temperature

mol = mole(s)

V = volume

V = volt(s)

atm = atmosphere

Part A

Directions: Every set of the given lettered choices below refers to the numbered statements or formulas immediately following it. Choose the one lettered choice that best fits each statement or formula and then fill in the corresponding oval on the answer sheet. Each choice may be used once, more than once, or not at all in each set.

Questions 1–6 refer to the choices in the following table:

Periodic Table (abbreviated)

1. The most electronegative element

2. The element with a possible oxidation number of –2

3. The element that would react in a 1 : 1 ratio with (D)

4. The element with the smallest ionic radius

5. The element with the smallest first ionization potential

6. The element with a complete p orbital as its outermost energy level

Questions 7–9 refer to the following terms.

(A) Reduction potential

(B) Ionization energy

(C) Electronegativity

(D) Heat of formation

(E) Activation energy

7. This is the energy change that accompanies the combining of elements in their natural states to form a compound.

8. This is the energy needed to remove an electron from a gaseous atom in its ground state.

9. This is the minimum energy needed for molecules to react and form compounds.

Questions 10–12 refer to the following heating curve for water:

10. In which part of the curve is the state of H2O only a solid?

11. Which part of the graph shows a phase change requiring the greatest amount of energy?

12. Where is the temperature of H2O changing at 4.18 J/g°C (or 1 cal/g°C)?

Questions 13–15 refer to the following diagram:

13. Indicates the activation energy of the forward reaction

14. Indicates the activation energy of the reverse reaction

15. Indicates the difference between the activation energies for the reverse and forward reactions and equals the energy change in the reaction

Questions 16–18 refer to the following elements in the ground state:

(A) Fe

(B) Au

(C) Na

(D) Ar

(E) U

16. A common metal element that resists reaction with acids

17. A monoatomic element with filled p orbitals

18. A transition element that occurs when the inner 3d orbital is partially filled

Questions 19 and 20 refer to the following:

(A) Radioactive isotope

(B) Monoclinic crystal

(C) Sulfur trioxide

(D) Sulfate salt

(E) Allotropic form

19. A substance that exhibits a resonance structure

20. A product formed from a base reacting with H2SO4

Questions 21–23 refer to the following terms:

(A) Dilute

(B) Concentrated

(C) Unsaturated

(D) Saturated

(E) Supersaturated

21. The condition, unrelated to quantities, that indicates that the rate going into solution is equal to the rate coming out of solution

22. The condition that exists when a water solution that has been at equilibrium and saturated is heated to a higher temperature with a higher solubility, but no additional solute is added

23. The descriptive term that indicates there is a large quantity of solute, compared with the amount of solvent, in a solution

Part B

ON THE ACTUAL CHEMISTRY TEST, THE FOLLOWING TYPE OF QUESTION MUST BE ANSWERED ON A SPECIAL SECTION (LABELED “CHEMISTRY”) AT THE LOWER LEFT-HAND CORNER OF YOUR ANSWER SHEET. THESE QUESTIONS WILL BE NUMBERED BEGINNING WITH 101 AND MUST BE ANSWERED ACCORDING TO THE FOLLOWING DIRECTIONS.

Directions: Every question below contains two statements, I in the left-hand column and II in the right-hand column. For each question, decide if statement I is true or false and if statement II is true or false and fill in the corresponding T or F ovals on your answer sheet. *Fill in oval CE only if statement II is a correct explanation of statement I.

Sample Answer Grid:

CHEMISTRY * Fill in oval CE only if II is a correct explanation of I.

I

II

101. Nonmetallic oxides are usually acid anhydrides

BECAUSE

nonmetallic oxides form acids when placed in water.

102. When HCl gas and NH3 gas come into contact,a white smoke forms

BECAUSE

NH3 and HCl react to form a white solid, ammonium chlorate.

103. The reaction of barium chloride and sodium sulfate does not go to completion

BECAUSE

the compound barium sulfate is formed as an insoluble precipitate.

104. When two elements react exothermically to form a compound, the compound should be relatively stable

BECAUSE

the release of energy from a combination reaction indicates that the compound formed is at a lower energy level than the reactants and thus relatively stable.

105. The ion of a nonmetallic atom is larger in radius than the atom

BECAUSE

when a nonmetallic ion is formed, it gains electrons in the outer orbital and thus increases the size of the electron cloud around the nucleus.

106. Oxidation and reduction occur together

BECAUSE

in redox reactions, electrons must be gained and lost.

107. Decreasing the atmospheric pressure on a pot of boiling water causes the water to stop boiling

BECAUSE

changes in pressure are directly related to the boiling point of water.

108. The reaction of hydrogen with oxygen to form water is an exothermic reaction

BECAUSE

water molecules have polar covalent bonds.

109. Atoms of different elements can have the same mass number

BECAUSE

the atoms of each element have a characteristic number of protons in the nucleus.

110. The transmutation decay of can be shown as

BECAUSE

the transmutation of is accompanied by the release of a beta particle.

111. and are isotopes of the element carbon

BECAUSE

isotopes of an element have the same number of protons in the nucleus but have a different number of neutrons.

112. The Cu2+ ion needs to be oxidized to form Cu metal

BECAUSE

oxidation is a gain of electrons.

113. The volume of a gas at 373 K and a pressure of 600 millimeters of mercury will be decreased at STP

BECAUSE

decreasing the temperature and increasing the pressure will cause the volume to decrease because

114. The pH of a 0.01 molar solution of HCl is 2

BECAUSE

dilute HCl dissociates into two essentially ionic particles.

115. Nuclear fusion on the sun converts hydrogen to helium with a release of energy

BECAUSE

some mass is converted to energy in a solar fusion.

116. The water molecule is polar

BECAUSE

the radius of an oxygen atom is greater than that of a hydrogen atom.

Part C

Directions: Every question or incomplete statement below is followed by five suggested answers or completions. Choose the one that is best in each case and then fill in the corresponding oval on the answer sheet.

49. Why was the CaCl2 tube placed between the generator and the tube containing the porcelain boat?

(A) To absorb evaporated HCl

(B) To absorb evaporated H2O

(C) To slow down the gases released

(D) To absorb the evaporated Zn particles

(E) To remove the initial air that passes through the tube

50. How many grams of hydrogen were used in the formation of the water that was a product?

(A) 1

(B) 2

(C) 4

(D) 8

(E) 9

51. What conclusion can you draw from this experiment?

(A) Hydrogen diffuses faster than oxygen.

(B) Hydrogen is lighter than oxygen.

(C) The molar mass of oxygen is 32 g/mol.

(D) Water is a triatomic molecule with polar characteristics.

(E) Water is formed from hydrogen and oxygen in a ratio of 1 : 8, by mass.

52. Which of the following is an observation rather than a conclusion?

(A) A substance is an acid if it changes litmus paper from blue to red.

(B) A gas is lighter than air if it escapes from a bottle left mouth upward.

(C) The gas H2 forms an explosive mixture with air.

(D) Air is mixed with hydrogen gas and ignited; it explodes.

(E) A liquid oil is immiscible with water because it separates into a layer above the water.

53. When HCl fumes and NH3 fumes are introduced into opposite ends of a long, dry glass tube, a white ring forms in the tube. Which statement explains this phenomenon?

(A) NH4Cl forms.

(B) HCl diffuses faster.

(C) NH3 diffuses faster.

(D) The ring occurs closer to the end into which HCl was introduced.

(E) The ring occurs in the very middle of the tube.

54. The correct formula for calcium hydrogen sulfate is

(A) CaH2SO4

(B) CaHSO4

(C) Ca(HSO4)2

(D) Ca2HSO4

(E) Ca2H2SO4

55. Forty grams of sodium hydroxide is dissolved in enough water to make 1 liter of solution. What is the molarity of the solution?

(A) 0.25 M

(B) 0.5 M

(C) 1 M

(D) 1.5 M

(E) 4 M

56. For a saturated solution of salt in water, which statement is true?

(A) All dissolving has stopped.

(B) Crystals begin to grow.

(C) An equilibrium has been established.

(D) Crystals of the solute will visibly continue to dissolve.

(E) The solute is exceeding its solubility.

57. In which of the following series is the pi bond present in the bonding structure?

I. Alkane

II. Alkene

III. Alkyne

(A) I only

(B) III only

(C) I and III only

(D) II and III only

(E) I, II, and III

Questions 58 and 59 refer to the following setup:

58. Why could you NOT use this setup for preparing H2 if the generator contained Zn + vinegar?

(A) Hydrogen would not be produced.

(B) The setup of the generator is improper.

(C) The generator must be heated with a burner.

(D) The delivery tube setup is wrong.

(E) The gas cannot be collected over water.

59. In a proper laboratory setup for collecting a gas by water displacement, which of these gases could NOT be collected over H2O because of its solubility?

(A) CO2

(B) NO

(C) O2

(D) NH3

(E) CH4

60. What is the approximate percentage of oxygen in the formula mass of Ca(NO3)2?

(A) 28

(B) 42

(C) 58

(D) 68

(E) 84

61. For the following reaction: N2O4(g) 2NO2(g),the Keq expression is

(A)

(B)

(C)

(D)

(E)

62. What is the Keq for the reaction in question 61 if at equilibrium the concentration of N2O4 is 4 × 10−2 mole/liter and that of NO2 is 2 × 10−2 mole/liter?

(A) 1 × 10−2

(B) 2 × 10−2

(C) 4 × 10−2

(D) 4 × 10−4

(E) 8 × 10−2

63. How much water, in liters, must be added to 0.50 liter of 6.0 M HCl to make the solution 2.0 M?

(A) 0.33

(B) 0.50

(C) 1.0

(D) 1.5

(E) 2.0

64. 4.0 grams of hydrogen are ignited with 4.0 grams of oxygen. How many grams of water can be formed?

(A) 0.50

(B) 2.5

(C) 4.5

(D) 8.0

(E) 36

65. Which structure is an ester?

(A)

(B)

(C)

(D)

(E)

66. What piece of apparatus can be used to introduce more liquid into a reaction and also serve as a pressure valve?

(A) Stopcock

(B) Pinchcock

(C) Thistle tube

(D) Flask

(E) Condenser

67. Which formulas could represent the empirical formula and the molecular formula of a given compound?

(A) CH2O and C4H6O4

(B) CHO and C6H12O6

(C) CH4 and C5H12

(D) CH2 and C3H6

(E) CO and CO2

68. The reaction Fe → Fe2+ + 2e− (+0.44 volt) would occur spontaneously with which of the following?

I. Pb → Pb2+ + 2e− (+0.13 volt)

II. Cu → Cu2+ + 2e− (-0.34 volt)

III. 2Ag+ + 2e− → 2Ag0 (+0.80 volt)

(A) I only

(B) III only

(C) I and III only

(D) II and III only

(E) I, II, and III

69. 2Na(s) + Cl2(g) → 2NaCl(s) + 822 kJHow much heat is released by the above reaction if 0.5 mole of sodium reacts completely with chlorine?

(A) 205 kJ

(B) 411 kJ

(C) 822 kJ

(D) 1,640 kJ

(E) 3,290 kJ

If you finish before one hour is up, you may go back to check your work or complete unanswered questions.

Answer Key

P R A C T I C E T E S T 1

1. D

14. B

104. T, T, CE

2. C

15. C

105. T, T, CE

3. B

16. B

106. T, T, CE

4. A

17. D

107. F, T

5. B

18. A

108. T, T

6. E

19. C

109. T, T

7. D

20. E

110. T, F

8. B

21. D

111. T, T, CE

9. E

22. C

112. F, F

10. A

23. B

113. T, T, CE

11. D

101. T, T, CE

114. T, T

12. C

102. T, F

115. T, T, CE

13. E

103. F, T

116. T, T

24. E

39. A

54. C

25. C

40. D

55. C

26. D

41. C

56. C

27. D

42. B

57. D

28. E

43. D

58. B

29. D

44. C

59. D

30. A

45. B

60. C

31. D

46. E

61. D

32. C

47. A

62. A

33. E

48. C

63. C

34. D

49. B

64. C

35. B

50. A

65. E

36. E

51. E

66. C

37. C

52. D

67. D

38. C

53. A

68. E

69. A

ANSWERS EXPLAINED

1.(D) The most electronegative element, F, would be found in the upper right corner of the table; the noble gases are exceptions at the far right.

2.(C) Elements in the group with (C) have a possible oxidation number of –2.

3.(B) Elements in the group with (B) react in a 1 : 1 ratio with elements in the group with (D), since one has an electron to lose and the other one needs an electron to complete its outer energy level.

4.(A) (A) loses 2 electrons to form an ion whose remaining electrons, being close to the nucleus, are pulled in closer because of the unbalanced 2+ charge.

5.(B) Since (B) has only one electron in the outer 4s orbital, it can more easily be removed than can an electron from the 3s orbital of (A), which is closer to the positive nucleus.

6.(E) All Group VIII elements have a complete p orbital as the outer energy level. This explains why these elements are “inert.”

7.(D) The heat of formation is the energy change caused by the difference in the initial energy and final energy of the system when elements in their standard state react to form a compound.

8.(B) The ionization energy is defined as the energy needed to remove an electron from the ground state of the isolated gaseous atom (or ion).

9.(E) The activation energy is defined as the minimum energy required for molecules to react . This is true for both exothermic and endothermic reactions.

10.(A) H2O is ice in Part A.

11.(D) H2O changes state at Parts B and D. The heat of vaporization at D (540 cal/g or 2.26 × 103 J/g) is greater than the heat of fusion (80 cal/g or 3.34 × 102 J/g) at B.

12.(C) Water is heating at 1°C/cal/g or 4.18 J/g/1°C in Part C.

13.(E) This is the energy needed to start the forward reaction.

14.(B) The part indicated by B represents the activation energy for the reverse reaction.

15.(C) The net energy released is the endothermic quantity indicated by C.

16.(B) Gold is known as a common noble metal because of its resistance to acids. Aqua regia, a mixture of HNO3 and HCl, will react with gold.

17.(D) Argon is the only element among the choices that is monoatomic in the ground state.

18.(A) Iron has 5 electrons in the d orbitals, which are partially filled.

21.(D) The condition described is the equilibrium that exists at saturation.

22.(C) With the increased temperature more solute may go into solution; therefore the solution is now unsaturated.

23.(B) The term “concentrated” means that there is a large amount of solute in the solvent.

101.(T, T, CE) Nonmetallic oxides are usually acid anhydrides, and they form acids in water.

102.(T, F) The white smoke formed is ammonium chloride, not ammonium chlorate.

103.(F, T) The reaction of barium chloride and sodium sulfate does go to completion since barium sulfate is a precipitate.

104.(T, T, CE) The product of an exothermic reaction is relatively stable because it is at a lower energy level than the reactants.

105.(T, T, CE) Both statements are true, and the reason explains the assertion.

106.(T, T, CE) Both statements are true, and the reason explains the assertion.

107.(F, T) The statement is false while the reason is true. Decreasing the pressure on a boiling pot will only cause the water to boil more vigorously.

108.(T, T) The statements are true, but the reason doesn’t explain the assertion.

109.(T, T) The statements are true, but the reason doesn’t explain the assertion.

110.(T, F) The transmutation equation is true and shows the release of an alpha particle, not a beta particle.

111.(T, T, CE) The two configurations of carbon are isotopes because they have the same atomic number but different mass numbers because has 6 protons and 7 neutrons while has 6 protons and 8 neutrons.

112.(F, F) Both the statement and the reason are false.

113.(T, T, CE) In going from 100°C to 0°C, the volume decreases as the gas gets colder; therefore, the temperature fraction expressed in kelvins must be to decrease the volume. To go from 600 mm to 760 mm of pressure increases the pressure, thus causing the gas to contract. The fraction must then be to cause the volume to decrease. You could use the formula

114.(T, T) Since HCl is a strong acid and ionizes completely in dilute solution of water (the [H+] and [H3O+] are the same thing), the molar concentration of a 0.01 molar solution is 1 × 10−2 mol/L.

pH = –log[H+] pH = –log[1 × 10–2] pH = –(–2)= 2

The pH is 2, but the reason, although true, does not explain the statement.

115.(T, T, CE) Both statements are true, and the reason explains the assertion.

116.(T, T) The water molecule is polar because its molecular structure has the more electronegative oxygen molecule at the one end and the two hydrogen molecules 105° apart. This causes the oxygen end to be more negatively charged than the hydrogen side of the molecule. The radius of the oxygen atom is greater than that of a hydrogen atom, but that has nothing to do with the polar nature of the water molecule.

24.(E) The total formula mass is:

25.(C) The catalyst, by definition, is not consumed in the reaction and ends up in its original form as one of the products. In this reaction, the catalyst is the MnO2.

26.(D) The ethyne molecule is the first member of the acetylene series with a general formula of CnH2n-2. It contains a triple bond between the two C atoms: H—CC—H.

27.(D) Heating molecules increases their kinetic energy.

28.(E) 3Ca + 2P + 8O = 13 atoms.

29.(D) Cesium and fluorine are from the most electropositive and electronegative portions, respectively, of the periodic chart, and thus form an ionic bond by cesium giving an electron to fluorine to form the respective ions.

30.(A) Dipole-dipole forces are due to the weak attraction of the nuclear positive charge of one atom to the negative electron field of an adjacent atom. They are much weaker than the others named.

31.(D) The molecular structure of water is that of a polar covalent compound with the hydrogens 105° apart.

32.(C) The most important considerations for a spontaneous reaction are (1) that the reaction is exothermic with a negative ΔH, so that once started it tends to continue on its own because of the energy released, and (2) that the reaction tends to go to the highest state of randomness, shown by a positive value for ΔS.

33.(E) Normal H+ acids and OH− bases form water and a salt (not necessarily a soluble salt).

34.(D) Every compound has a charge of 0. H usually has +1, and O usually has -2, so

36.(E) HCl and H3O+ give acid solutions, as does CuSO4 (the salt of a weak base and a strong acid) when it hydrolyzes in water. I, II, and III are correct.

37.(C) Mass is a constant and is not dependent on position or surrounding conditions.

38.(C) The complete outer energy levels of electrons of the smallest noble gases have the highest ionization potential.

39.(A) Only NO and NO2 fit the definition of the Law of Multiple Proportions, in which one substance stays the same and the other varies in units of whole integers.

40.(D) Increasing the concentration of one or both of the reactants and removing some of the product formed would cause the forward reaction to increase in rate to try to regain the equilibrium condition. II and III are correct.

41.(C) I and II are correct. The equation is 2Na + 2H2O → 2NaOH + H2(g). The coefficients include a 1 and a 2.

43.(D) H2SO4 + 2NaOH → 2H2O + Na2SO4 is the equation for this reaction. 1 mol H2SO4 = 98 g. 1 mol NaOH = 40 g. Then 49 g of H2SO4 = mol H2SO4. The equation shows that 1 mol of H2SO4 reacts with 2 mol of NaOH for a ratio of 1 : 2. Therefore, mol of H2SO4 reacts with 1 mol of NaOH in this reaction. Since 1 mol of NaOH equals 40. g, and 80. g of NaOH is given, 40. g of it will remain after the reaction has gone to completion.

45.(B) If 1.43 g is the mass of 1 L, then the mass of 22.4 L, which is the gram-molecular volume of a gas at STP, will give the gram-molecular mass. Then, 22.4 L × 1.43 g/L = 32.0 g, the gram-molecular mass.

47.(A) Gibbs free energy combines the overall energy changes and the entropy change. The formula is ΔG = ΔH – TΔS. Only if ΔG is negative will the reaction be spontaneous in the forward direction.

48.(C) The CuO was reduced while the H2 was oxidized, forming H2O + Cu. The reduction and oxidation reaction is:

CuO(s) + H2(g) → H2O(l) + Cu(s)

49.(B) The purpose of this CaCl2 tube is to absorb any water evaporated from the generator. If it were not present, some water vapor would pass through to the final drying tube and cause the weight of water gained there to be larger than it should be from the reaction only.

50.(A) Since 8 g of O2 was lost by the CuO, the weight of water gained by the U tube came from 8 g of O2 and 1 g of H2, to make 9 g of water that it absorbed.

51.(E) The weight ratio of water is 1 g of H2 to 8 g of O2 or 1 : 8. The other statements are true but are not conclusions from this experiment.

52.(D) This is the only observation of the group. All the others are conclusions. Remember that an observation is only what you see, smell, taste, or measure with a piece of equipment.

53.(A) The phenomenon is the formation of the white ring, which is NH4Cl. Although the distance traveled by each gas could be measured to verify Graham’s Law of Gaseous Diffusion, this was not asked. The relationship is that the diffusion rate is inversely proportional to the square root of the gas’s molecular weight.

54.(C) Since Ca2+ and HSO4− combine, the formula is Ca(HSO4)2.

55.(C) NaOH is 40 g/mol. 40 g in 1 L makes a 1 M solution.

56.(C) A saturated solution represents a condition where the solute is going into solution as rapidly as some solute is coming out of the solution.

57.(D) II and III are correct since the double-bonded carbons in the alkene and the triple-bonded carbons in the alkyne series have pi bonds.

58.(B) The thistle tube is not below the level of the liquid in the generator and the gas would escape into the air. (Vinegar is an acid and would produce hydrogen.)

59.(D) NH3 is very soluble and could not be collected in this manner. All others are not sufficiently soluble to hamper this method of collection.

64.(C) The reaction equation and information given can be set up like this:

Given

Given

4.0 g

4.0 g

xg

2H2

+

O2

→

2H2O

4.0 g

32 g

36g

Studying this shows that the limiting element will be the 4 g of oxygen since 4 g of H2 would require 32 g of O2. The solution setup is

65.(E) The functional group of an ester is . This appears only in (E).

66.(C) The thistle tube serves both these purposes.

67.(D) The empirical formula is a representation of the elements in their simplest ratio. Therefore, CH2 is the simplest ratio of the molecular formula C3H6.

68.(E) I, II, and III would occur since their reduction reactions would be –0.13, +0.34, +0.80 V, respectively. These numbers added to +0.44 V separately give a positive E0 for the reaction with Fe.

69.(A) In the equation, 2 mol of Na release 822 kJ of heat. If only 0.5 mol of Na is consumed, only one-fourth as much heat will be released: × 822 = 205 kJ.

CALCULATING YOUR SCORE

Your score on practice Test 1 can now be computed manually. The actual test is scored by machine, but the same method is used to arrive at the raw score. You get one point for each correct answer. For each wrong answer, you lose one-fourth of a point. Questions that you omit or that have more than one answer are not counted. On your answer sheet mark all correct answers with a “C” and all incorrect answers with an “X”.

Determining Your Raw Test Score

Total the number of correct answers you have recorded on your answer sheet. It should be the same as the total of all the numbers you place in the block in the lower left corner of each area of the Subject Area summary in the next section.

A. Enter the total number of correct answers here: ________ Now count the number of wrong answers you recorded on your answer sheet.

B. Enter the total number of wrong answers here: ________ Multiply the number of wrong answers in B by 0.25.

C. Enter that product here: ________ Subtract the result in C from the total number of right answers in A.

D. Enter the result of your subtraction here: ________

E. Round the result in D to the nearest whole number: ________.This is your raw test score.

Conversion of Raw Scores to Scaled Scores

Your raw score is converted by the College Board into a scaled score. The College Board scores range from 200 to 800. This conversion is done to ensure that a score earned on any edition of a particular SAT Subject Test in Chemistry is comparable to the same scaled score earned on any other edition of the same test. Because some editions of the tests may be slightly easier or more difficult than others, scaled scores are adjusted so that they indicate the same level of performance regardless of the edition of the test taken and the ability of the group that takes it. Consequently, a specific raw score on one edition of a particular test will not necessarily translate to the same scaled score on another edition of the same test.

Because the practice tests in this book have no large population of scores with which they can be scaled, scaled scores cannot be determined.

Results from previous SAT Chemistry tests appear to indicate that the conversion of raw scores to scaled scores GENERALLY follows this pattern:

Note that this scale provides only a general idea of what a raw score may translate into on a scaled score range of 800–200. Scaling on every test is usually slightly different. Some students who had taken the SAT Subject Test in Chemistry after using this book had reported that they have scored slightly higher on the SAT test than on the practice tests in this book. They all reported that preparing well for the test paid off in a better score!

DIAGNOSING YOUR NEEDS

After taking Practice Test 1, check your answers against the correct ones. Then fill in the chart below.

In the space under each question number, place a check if you answered that question correctly.

EXAMPLE:

If your answer to question 5 was correct, place a check in the appropriate box.

Next, total the check marks for each section and insert the number in the designated block. Now do the arithmetic indicated, and insert your percent for each area.

* The subject areas have been expanded to identify specific areas in the text.

* The subject areas have been expanded to identify specific areas in the text.

Answer Sheet

P R A C T I C E T E S T 2

Determine the correct answer for each question. Then, using a No. 2 pencil, blacken completely the oval containing the letter of your choice.